A small electric immersion heater is used to heat 75g of water. Heater is labele
ID: 1481490 • Letter: A
Question
A small electric immersion heater is used to heat 75g of water. Heater is labeled 50 watts. Calculate time required to bring all water from 20 degree to 100 degree Celsius and evaporate it, ignoring heat loss.( specific heat = 4186 latent heat = 2264.76) A small electric immersion heater is used to heat 75g of water. Heater is labeled 50 watts. Calculate time required to bring all water from 20 degree to 100 degree Celsius and evaporate it, ignoring heat loss.( specific heat = 4186 latent heat = 2264.76)Explanation / Answer
Mass of water m = 75 g = 75 x10 -3 kg
Heat required to evaporate 75 g of water from 20 o C is Q = mC dt + mL
Where C = Specific heat of water = 4186 J / kg o C
L = latent heat of vaporisation of water = 2264.76 x10 3 J / kg
dt = temprature difference = 100 -20 = 80 o C
Substitute values you get Q = (75 x10 -3 x4186x80) +(75x10 -3 x 2264.76 x10 3 )
= 25116 J + 169857 J
= 194973 J
Power of heater P = 50 watts
Required time t = Q / P
= 194973 J / 50 watts
= 3899.46 s
= (3899.46 / 3600) hr
= 1.083 hr